Circular motion 5 - WordPress.com · 2018. 2. 28. · 2Prof. Muhammad Amin Chapter-5 circular...

Prof. Muhammad Amin 1

Chapter-5 circular motion

Type equation here.

Q. 5.1 Define and explain centripetal force and derive relation for it?

Ans.

Centripetal Force

“The force required to bend the normally straight path into circular path is called

centripetal or radial force. Its direction is always towards the centre of a circle.”

It is denoted by cF

Explanation

Let us take an example of a body attached to a string whirled on circular path as shown

in figure. Observations shows that if string is snapped it will not continue its motion on

circular path rather it will follow straight path.

Hence centripetal force must be furnished to move the body on circular path

Relation for Centripetal force

Consider a particle that moves from point A to B with uniform speed v on a circular path of

radius “r”. If ∆v is the change in velocity then acceleration is given by

𝑎 =∆v

∆𝑡 − − − − − (1)

Where ∆t the time is required by the particle to move from A to B. let v1⃗⃗ ⃗ and v2⃗⃗ ⃗ are velocities

at point A and B respectively as shown in figure (a)

If S is the distance covered by the particle from point A to b then

∆𝑡 =S

V

Putting this value in equation (1)

𝑎 =∆v

𝑆 (v) − − − − − (2)

Now we draw a triangle ∆DOE as shown in figure (b)

Such that

OD⃗⃗ ⃗⃗ ⃗ is parallel and equal to v1⃗⃗ ⃗

OE⃗⃗⃗⃗ ⃗ is parallel and equal to v2⃗⃗ ⃗

Circular motion Chapter

5

Important long questions



As radius of the circle is perpendicular to its tangent, so v1⃗⃗ ⃗ and v2 ⃗⃗⃗⃗ ⃗ are perpendicular to OA

and OB.

⇒ < 𝐴𝑂𝐵 = < 𝐷𝑂𝐸

Moreover v1 = v2 = v and OA =OB

Hence for Isosceles triangles AOB and DOE , we can write

DE

DO=

AB

OB

∆v

v=

AB

r− − − − − −(3)

When ∆t → 0 then AB ≈ S

Equation (3) becomes

∆v =S

r (v)

Putting this value in equation (2)

𝑎 =

Sr (v)

𝑆 (v)

2

__________(4)c

va

r

Equation (4) represents centripetal acceleration.

In term of angular velocity, V = r

2 2 2

c

r ra

r r

2 __________(5)cor a r

Now

According to Newton’s second law of motion, the centripetal force is given by

______________ 6___c cF m a

Putting the value of ac from equation (4)

2

________________ 7_c

VF m

r

In term of angular velocity, using equation (5) in equation (6)

2 r ______________ 8___cF m

Equations 7 and 8 represents centripetal acceleration.



Q. 5.2 what is meant by moment of inertia. Explain its physical significance.

Ans. Moment of Inertia

“The analogous of mass in rotation motion is called moment of inertia”

OR

Moment of inertia of a particle is “the product of its mass and square of the

perpendicular distance from axis of rotation”

Mathematically

𝐼 = 𝑚𝑟2

Its S.I unit is kg m2

Explanation

Consider a mass attached to a light rod, it can rotate about the

point ‘o’ as shown in figure

The mass of the rod is negligible.A force F is acting on the

mass perpendicular to the road and hence, this will accelerate

the mass according to second law of motion

a _________________ 1c TF m

This force will rotate the mass about ‘o’. Tangential acceleration ‘aT’ is related to

angular acceleration ‘’ by the equation

Ta r Putting in (1)

__________(2)F mr

Multiplying by r

2rF mr

Thus 2rF mr I T is analogue of the Newton’s second law of motion F = ma.

Moment of inertia is defined as “the product of mass of the particle and square of its

perpendicular distance from the axis of rotation.”

i.e. I = m r2

Moment of inertia of a rigid body

A body is made of n – small particles of masses m1, m2, m3,……………… and if the

distances of the particles from the axis of rotation ‘o’ be r 1,r2,……. rn,



Then moment of inertial of the body is.

2 2 21 1 2 2 ....... n nI m r m r m r

2

1

i n

i ii

or I m r

Physical Significance

For a body in linear motion, the acceleration is directly proportional to the force acting

upon the body.

. . ( tan )F

i e F ma or m Cons ta

Where m is the mass of a moving body which is direct measure of inertia similarly for a

rotating body, the angular acceleration is directly proportional to the torque acting on it

or the ration of the torque to the angular acceleration is constant.

tanT

Cons t

So moment of inert of a body is analogous to the mass in linear motion. Hence

moment of inertia may be described as rotational mass of a body.

Q. 5.3 (a) Define rotational kinetic energy and derive a relation for it?

(b) Discuss rotational kinetic energy of a disc and a hop. Also derive the relation

for the velocities of a disc and hoop start moving down an inclined plane.

Ans. Rotational Kinetic Energy

The energy due to the spinning of a body about an axis is called rotational kinetic

energy.

Explanation

Consider a body is spinning about an axis with constant angular velocity . Everybody

is made up of large number of tiny pieces. Each piece of the body is moving in a

circular path and therefore, has some kinetic energy. The sum of the kinetic energies of

these pieces will give the kinetic energy of the whole body.If a piece of mass m i is at a

distance r i from the axis of rotation as shown in fig.

The linear speed of the piece is given by

i iV r

The kinetic energy of this piece is

21. .

2i ii

K E m V

21

. .2

i iiK E m r

2 21

2i im r

Let us suppose that the tiny pieces of the body having masses m1, m2, m3………… at

distances r1, r2, r3…….



The K.E of the first piece 2 22 1

1

2m r

The K.E of the second piece 2 22 2

1

2m r

Total K.E 2 2 22 1 2 2

1.....

2m r m r

Where 2 21 1 2 2 .....m r m r I

I = moment of inertia of the body about the axis of rotation.

2

.

1. . . .

2rotTotal rotational K E K E I

(b) Rotational Kinetic Energy of a Disc And Hoop

Let us use the idea of rotational K.E to compare the velocity with which a disc and a

hoop reach the bottom of an inclined plane.

Figure shows that disc and the hoop start moving down an inclined plane of height h,

their motion consists of rotational and translational motions.

If no energy is lost in friction, the total K.E of the disc or hoop on reaching the bottom

of the incline must be equal to its potential energy at the top P.E at the top = rotational

K.E.at bottom + translational K.E at the bottom.

2 21 1____________(1)

2 2mgh I mV

Where and V are the angular and linear speeds at the bottom m is the mass of the

rolling body.

For Disc

21,

2I mr Putting in equation (1)

Where r is its radius.

2 2 21 1 1

2 2 2mgh mr mV

2 2But V r

2 2 2or V r

2 21 1 1

2 2 2mgh mV V

or m gh m 2 1 1

4 2V

2 3

4gh V



2 4 4

3 3

gh ghV or V

This is the speed of disc on reaching the bottom of inclined plane

For Hoop

2I m r

Putting in equation (1)

2 2 21 1

2 2mgh mr mV

2 2 2V r

2 21 1

2 2mgh mV mV

mgh m 2V

2or V gh

or V gh

This is the speed of hoop on reaching the bottom of inclined plane

Q. 5.4 An object is suspended from a spring balance by a string and spring balance in a

lift. Find the apparent weight of the object in the following cases.

(a) When the lift is at rest.

(b) When the lift moves up with uniform acceleration.

(c) When the lift moves down with uniform acceleration.

(d) When the lift falling freely under the action of gravity.

Ans. Real weight

The force with which an object is attracted towards the center of earth is called its real

weight i.e W = mg

Apparent Weight

The force which prevents a body from falling in a certain frame of reference is known

as its apparent weight

Example of an elevator

Consider an object of mass m is suspended from a spring balance

by means of a string and spring balance is attached to the ceiling

of a lift as shown in figure.

The tension in the string is always equal in magnitude to the

weight of the object.

The reading of the spring balance i.e. tension T will

represent the apparent weight of the object.



(a) When the lift is at rest

According to Newton’s second law of motion, the resultant force

on the object is zero.

So a = o

T ma

T m o

T o

________________(1)T

From equation (1), it is found that the apparent weight of the object is equal to its real

weight.

(b) When the lift moves up with uniform acceleration.

Let the lift moves upwards with an acceleration a.

T

T – = F (resultant force)

,mg F ma

T mg ma

____________(2)or T mg ma

This equation shows that the apparent weight of the object has increased by an amount

ma.

(c) When the lift moves down with uniform acceleration.

Let the lift moves downwards with an acceleration ‘a’ as shown in fig (b).

T

–T = F (resultant force)

,mg F ma

mg T ma

____________(3)or T mg ma

Equation (3) shows that the apparent weight of the object has

decreased by an amount ma.

(d) When the lift falling freely under the action of gravity.

If the cable supporting the lift breaks, the lift will fall down like

a free body with an acceleration equal to g.

, (3)a g Putting in equation

T mg mg

or T o

Hence the spring balance will read zero weight and the man in the lift will think that

the object has no weight i.e. it is in the state of weightlessness.

at rest

a=0

T=w

Acceleration downward

w-T=ma

T=w-ma

Acceleration upward

T-W=ma

T=w+ma



Q. 5.5 Write short note on artificial gravity?

Ans. Artificial Gravity

Artificial gravity is produced by rotating a space craft about its own axis.

Explanation

In satellite all the objects become weightless. This produces a lot of difficulties for

astronauts in doing experiments. This problem is removed by producing artificial

gravity in the satellite. Rotating the satellite about its own axis produces the artificial

gravity.

Expression for Frequency

Consider a ring shaped space craft as shown in fig.

The outer radius of the space craft is R and it rotates around its own axis with angular

speed ω. At any point on the outer rim we have a centripetal acceleration is given by

the formula.

2

________________(1)c

Va V

R

Where V = linear speed of the space croft

As V = Rω Putting in equation (1)

2 22 ________(2)c

Ra R

R

If time period of space craft is T, then

2

TPutting in equation (2)

2 2

2

2 4ca R R

T T

2

2

14 _________(3)cor a R

T

Frequency can be expressed as

1f

TPutting in equation (3)

2 24ca R f

2

2

1

4 2c ca a

f or fR R

As force of gravity provides the centripetal acceleration, therefore, a c = g.

1

2

gf

R

When the space craft rotates with this frequency, the artificial gravity is produced in it.

Q. 5.6 What are geostationary orbits? Calculate the value of the radius measured from

the centre of the earth for a geostationary satellite?

Ans. Geostationary Orbit

A geo-stationary orbit is that in which the period of rotation of the satellite around the

earth is exactly equal to the period of rotation of the earth about its axis. A satellite



with an orbital period of 24 hours will always be at the point above the earth’s surface.

Such satellites are called geostationary satellites or geosynchronous satellites.

They remain at a certain height 3600 Km above the equator of the earth and move in the

same direction as the earth is rotating.

Uses

Geostationary satellites are useful for worldwide communication. Weather observations

navigation and military purpose

Radius of Geostationary Orbit

The orbital speed of satellite is given by

___________(1)GM

Vr

Where

M = mass of the earth

G = gravitational constant.

r = radius of the orbit

This orbital speed is equal to the average speed of the satellite in one

2. . ______(2)

S ri e V

t T

Where T is the period of revolution of the satellite which is equal to one day

As the earth completes its one revolution in one day and the satellite will revolve

around the earth in one day, so the satellite at position A will always stay over the same

position A on the earth as shown in fig.

By comparing equation (1) and (2),

2 r GMwe have

T r

Taking square on both sides

2 2

2

4 r GM

T r

2 2 24 r r GM T

23

24

GMTor r

T = period of the earth = period of the satellite.

T = 234 hours

T = 24 60 60 S = 86 400 s

G = 6.67 10–11 N–m2 / Kg2

M = 6 1024 Kg

211 243

2

6.67 10 6 10 86400

4 3.14r

or r = (7.575 1022)1/3 = 4.23 107 m

or r = 4.23 104 Km = 42300 Km



Which is the radius of a geostationary satellite from the centre of the earth.

If h is the height of the satellite above a particular point on the surface of the earth,

then

R = R + h

Or h = r – R

Radius of the earth = R = 6400 Km

h = 423 100 – 6400

h = 35900 Km 36000 Km



5.1 Explain the difference between tangential velocity and the angular velocity. If one

of these is given for a wheel of known radius, how will you find the other?

Ans:

TANGENTIAL VELOCITY ANGUALR VELOCITY

Velocity of a body along the

tangent is known as tangential

velocity of linear velocity.

Its unit is m/s

Its direction is along tangent

,t

dv v r

t

Angular velocity of a body is the

rate of change of angular

displacement.

Its unit is radian/sec

Its direction is along the axis of

rotation.

v

t r

If one of them is given for a wheel of known radius, then other can be calculated using

the relation v = r

5.2 Explain what is meant by centripetal force and why it must be furnished to an

object if the object is to follow a circular path?

Ans: Definition

“The force needed to bend the normally straight path of the particle into a circular path

is called centripetal force”

Centripetal Force only changes the direction of motion

When a force acts perpendicular to the direction of motion of a

body then that force changes only the direction of motion of the

body. When a body moves in a circular path then at every instant

its direction of motion changes.

Hence it must be furnished so that direction of motion of body

changes continuously due to which straight path bends into

circular path.

5.3 What is meant by moment of inertia? Explain its significance.

Ans: Definition

“The rotational analogous of linear mass is called moment of inertia”

OR

Moment of inertia of a particle is “the product of its mass and square of the

perpendicular distance from axis of rotation”

Mathematically

𝐼 = 𝑚𝑟2

Its S.I unit is kg m2

Important short questions



Significance

It plays same role in angular motion which inertia plays in linear motion. It resists

angular / circular motion as inertia resists linear motion.

It may be noted that moment of inertia depends not only on mass m but also on r 2.

5.4 What is meant by angular momentum? Explain the law of conservation of angular

momentum.

Ans: Angular Momentum

The moment of linear momentum of a body about a point is called angular momentum.

OR

The cross product of position vector r and linear momentum P of an object is called

angular momentum. It is denoted by L .

Mathematically, L r p

Its S.I unit kg m2 s-

Law of conservation of angular momentum:

It states that if no external torque acts on a system, the total angular momentum of the

system remains constant.

1 2

1 1 2 2

constant

I =I constant (For an isolated system)

TotalL L L

5.5 Show that orbital angular momentum L0 = mvr or L = I𝝎

Ans: Proof

Consider a boy of mass m is moving in a circular path of radius r with speed V, the

angular momentum of the body is given by

oL r P

The magnitude of angular momentum is given by

0

0

0

0

sin

( 90 )

sin 90

( sin 90 1)

,

,

o

o

o

L rp

L rp

L rp

As p mv

So

L rmv

Hence, Proved, L0 = mvr

Also 𝑣 = 𝑟 𝜔

⇒ L0 = m ( r ω) r = m r2 𝜔

L0= I 𝜔

5.6 Describe what should be the minimum velocity, for a satellite, to orbit close to the

Earth around it.

Ans: Definition

“The minimum velocity necessary to put the satellite into an orbit close to the surface

of Earth is called critical velocity.”



It is given by the relation.

69.8 6.4 10v gR

3 17.9 10 secv m

So, v = 7.9kms-1

5.7 State the direction of the following vectors in simple situation; angular momentum

and angular velocity.

Ans: The direction of angular momentum and angular velocity is determined by right hand

rule.

For Angular Momentum

We know that:

L r p

This shows that the direction of angular momentum is perpendicular to the plane

containing r and p .

In case of circular motion, angular momentum is perpendicular to the plane of circle

and is along axis of rotation.

For Angular velocity

The direction of angular velocity is perpendicular to the plane of the circle and is along

the axis of rotation.

5.8 Explain why an object, orbiting the Earth, is said to be freely falling. Use your

explanation to point out why objects appear weightless under certain

circumstances.

Ans: When an object is orbiting around the Earth, force of gravity provides the necessary

centripetal force, as no force is holding that object. Therefore, it is freely falling frame

of reference. The curvature of path of the object and the Earth prevent it from falling

on the surface of Earth.

In a freely falling frame of reference, the frame of reference and the objects inside the

frame of reference are falling at the same rate i.e. same acceleration a = g, therefore

they appear to be weightless.

5.9 When mud flies off the tyre of a moving bicycle, in what direction does it fly?

Explanation

Ans: When tyre of bicycle moves, then init ially mud rotates with the tyre. The adhesive

force between mud and the tyre provides necessary centripetal force to rotate the mud

along with tyre.

When speed of tyre increases, more centripetal force is required by mud to rotate with

tyre. Since, adhesive force is not enough to provide necessary centripetal force.

Therefore, the mud flies off the tyre in a direction of tangent to the circular path.



5.10 Why does a diver change his body positions before and after diving in the pool?

Ans:

Diver changes his bosy positions before and after the diving of the pool for the

effective use of law of conservation of momentum.

i.e

𝐿 = 𝐼𝜔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

When a diver jumps off the diving board, he has small angular velocity about the

horizontal axis through his centre of gravity. When he draws his legs and arms close to

his body, his moment of inertia decreases and to conserve angular momentum, his

angular velocity increases and he spins faster. This enables the diver to take extra

somersaults.

5.11 Explain how many minimum number of geo-stationary satellites are required for

global coverage of T.V transmission .

Ans: To cover the whole earth, three properly positioned geo stationary satellites are

required. Each satellite covers an angle of 120 0 longitude and three satellites will be

covering complete angle of 3600.

5.12 What is angular displacement? How can we determine its direction?

Ans. Angular Displacement

It is the angle subtended at the centre of a circle by a particle moving

along the circumference in a given time.

Direction

Its direction can be determined by right hand rule.

Grasp the axis of rotation in right hand with fingers curling in the

direction of rotation the thumb points in the direction of angular

displacement.

5.13 Prove that S r

Ans. Consider an arc of length s of a circle of radius r which subtends an angle at the

centre of circle. By the definition of radian, its value in radian is given as

Arc length

radradius

S

r

s r



5.14 Show that I

Ans. Let the rigid body is made up of n small pieces of masses m1, m2,…i….mn at distances

r1, r2, ………….rn from the axis of rotation. If rigid body is rotating, then the magnitude of

torque acting on m1, is

2

1 1 1m r

Similarly torque on m2 is

2

2 2 2m r

2 2 2

1 1 2 2 ..........Total n nm r m r m r

2

1

n

i i

i

m r I

I

5.15 Convert kgm2s-1 into JS.

Ans. 2 1 skgm s

s

= 2

2

.kgm s

s

= 2

. .kgm m s

s

As 2/kg m s N

= N. m .s

As N.m = J

Hence

Kgm2s-1 = Js

5.16 Why does the coasting rotating system slow down as water drips into the beaker?

Ans. When water drips into the beaker, the mass of the contents in the beaker increases

which increases the moment of inertia. So the angular velocity decreases according to law

of conservation of momentum.

5.17 What will be effect on the speed of satellite by increasing its mass?

Ans. We know that GM

vr

This relation shows that speed of satellite is independent of the mass of the satellite.

Therefore change of mass will not affect its speed.

5.18 What is meant by INTELSAT?

Ans. It means international telecommunication satellite organization. It operates at

microwaves frequencies of 4, 6, 11 and 14 GHz and has a capacity of 30, 000 two way

telephone circuits plus three TV channels.

5.19 Differentiate between Real and apparent weight of a body.

Ans.

Real Weight Apparent Weight

(i) It means. gravitational pull of

earth on that object.

(ii) It is measured as W = mg

The force which prevent the body from freely

falling is called apparent weight.

It is measured as tension in the spring balance.



5.20 If a person is falling in an elevator freely. What will be his weight? Measured by

himself.

Ans. The force required to prevent a body from free fall is called its apparent weight. If a

person is falling freely in an elevator, its apparent weight is zero. Therefore he will be

in weightless. It means. that a freely falling body in an elevator feels no w eight by

himself.

Since apparent weight in descending lift is given by

T = mg – ma

But g = a = 9.8m/s2

T = mg – mg

= 0 N

5.21 Find the velocity of a disc rolls down along an inclined plane of height 10m?

Ans. h = 10m

v = ?

v = 4

3

gh

= 4 9.8 10

3 10

v = 392

3

v = 11.43 m/s

5.22 Describe the uses of geostationary satellite?

Ans. Uses of geostationary satellites are

(i) Worldwide communication

(ii) Weather observations

(iii) Navigation

(iv) Other military uses

5.23 Is any work done by centripetal force?

Ans. No, in circular motion, centripetal force is always acting perpendicular to the

displacement of the moving body

.w F d

= F d cosθ θ = 90o

= F d cos 90o

w = 0

Hence, work done by the centripetal force is zero.

5.24 Prove that 1 Radian = 57.3o

Ans. If OP is rotating, the point P covers a distance S = 2πr in one revolution. P.

In radian it would be 2s r

r r

= 2π

1 Revolution = 2π rad = 360o

1rad = 360

2

o

d

Fc

0

P

1P

S



1rad = 57.3o

5.25 Show that v = r ω

Ans. Suppose during the course of its motion, the point P moves

through a distance P1P2 = ∆s in a time interval ∆t during

which reference line OP has an angular displacement ∆θ

radian during this interval. ∆s and ∆θ are related by Eq.

S r

Dividing both sides by ∆t

S

rt t

………..(i)

In the limit when 0t the ratio /S t represents v, the magnitude of the velocity

with which point P is moving on the circumference of the circle. Similarly / t

represents the angular velocity of the reference line OP. So equation (i) becomes.

v r

NUMERICAL PROBLEMS

4.1 A tiny laser beam is directed from the Earth to the Moon. If the beam is to have a diameter

of 2.50 m at the Moon, how small must divergence angle be for the beam? The distance of

Moon from the Earth is 3.8 x 108m.

Solution:

Data: Diameter of beam = length of arc = S = 2.50 m

Distance of Moon from the Earth = radius of circular arc = r = 3.8 x 108m

To find: Divergence angle = = ?

Calculation:

Using the formula

S = r

or = S/r

Putting the values, we get

9

8

2.50θ 6.6 10

3.8 10

rad

= 6.6 x 10-9 rad Ans

4.1 A gramophone record turntable accelerates from rest to an angular velocity of 45.0 rev min-1

in 1.60s. What is its average angular acceleration?

Solution:

Data: Initial angular velocity = iω =0

final angular velocity = fω =45.0 rev min-1

45.0×2π

= =1.5π60

rads-1

Time = t = 1.60s

To Find: Average angular acceleration = = ?

Calculation:

f iω -ωα=

t


Numerical Problems



1.5π-0

α= =2.951.60

rads-2

α = 2.95rad s-2 Ans

4.1 A body of moment of inertia I = 0.80 kg m2 about a fixed axis, rotates with a constant

angular velocity of 100 rad s-1. Calculate its angular momentum L and the torque to sustain

this motion.

Solution:

Data: Moment of inertia = I = 0.80 Kgm2

Angular velocity = = 100 rads-1

To Find: Angular momentum = L = ?

Torque = = ? Calculation: L = I Putting the values, we have L = 0.80 x 100 = 80 kgm2s-1

or L = 80 kgm2s-1 x s

s

= 80 kgm2s-2 s L = 80 Js Ans. Torque can be found by using the relation τ=Iα

Since the body has constant angular velocity, therefore angular acceleration is zero. Thus,

τ= I×0=0

τ = 0 Ans

4.1 Consider the rotating cylinder as shown in Fig. Suppose that m = 5.0 kg, F = 0.60 N and r = 0.20m. Calculate (a) the torque acting on the cylinder, (b) the angular acceleration of the

cylinder. (Moment of inertia of cylinder = 1

2mr2)

Solution: Data: Mass of cylinder = m = 5.0 kg Force acting on cylinder = F = 0.60 N Distance = r = 0.20m To Find: Torque acting on cylinder = = ?

Angular acceleration of cylinder = a = ? Calculation:

Using the formula of torque τ= r Fsinθ

r and F are perpendicular to each other. τ=r Fsin90


τ = 0.20×0.60×1

.τ = 0.12Nm Ans

Now using the formula τ = Iα

Where I is the moment of inertia of cylinder which is equal to 21mr

2

Therefore,

21τ = mr α

2

Putting the values, we have

21τ= ×5.0×(0.20) α

2

0.12 = 0.1

or -2α=0.12/0.1=1.2rads

-2α = 1.2rads Ans

4.1 Calculate the angular momentum of a star of mass 2.0 x 1030 kg and radius 7.0 x 105 km.



If it makes one complete rotation about its axis once in 20 days, what is its kinetic energy?

Solution:

Data: Mass of a star = M = 2.0 x 1030

Radius of star = R = 7.0 x 105 km

= 7.0 x 108 m

Time period of star = T = 20 days

= 20 x 24 x 60 x 60s

= 1728000s

To find: Angular momentum of star = L = ?

Kinetic energy = K. E. = ?

Calculation:

Using the formula

L = I

Where 2

I=5

MR2 for a sphere

And 2π

ω=T

Thus, 22 2πL= MR ×

5 T

Putting the values, we have

30 8 22×2×10 ×(7×10 ) ×2×3.14

L =5×1728000

46

3

2×2×7×7×2×3.14×10=

5×1728×10

43=0.142462963×10 42 -1 -1 -1L =1.4×10 kgms ×s /s

42L =1.4×10 Js Ans.

(b) Using the formula of kinetic energy K.E. = 2Iω

2

Putting the values of I and , we have

K.E. = 2 21 2× ×MR ×(2π/T)

2 5

or K.E. = 30 8 2 21 2× ×2×10 (7×10 ) (2π/20×24×3600)

2 5

30 16 -111= ×2×10 ×49×10 ×(1.3×10 )

5

35= 25.48×10 J

35= 25×10 ×10/10

K.E. = 2.5x 1036J Ans.

4.1 A 1000 kg car traveling with a speed of 144 km h-1 round a curve of radius 100 m. Find the

necessary centripetal force.

Solution:

Data: Mass of the car = M = 1000 kg

Speed of car = v = 144 kmh-1

-1144×1000= =40ms

60×60

Radius of the curve = r = 100m



To Find: Centripetal force = F = ?

Calculation:

Using the formula 2

c

mvF =

r

Putting the values, we get 2

4

c

1000×(40)F = =1.60×10 N

100

Fc = 1.60 x 104 N Ans

4.1 What is the least speed at which an aeroplane can execute a vertical loop of 1.0 km radius

so that there will be no tendency for the pilot to fall down at the highest point?

Solution:

Data: Radius of loop = r = 1.0 km = 1000 m

Acceleration due to gravity = g = 9.8 ms-2

To find: Speed of aeroplane = v = ?

Calculation:

When an aeroplane executes a circular loop, the centripetal acceleration is supplied by gravity and

we have

or

2

c

2

a =g=v /r

v =rg

v= rg


v = 1000 x 9.8

v = 99 ms-1 Ans.

4.1 The Moon orbits the Earth so that the same side always faces the Earth. Determine the ratio

of its spin angular momentum (about its own axis) and its orbital angular momentum. (In

this case, treat the Moon as a particle orbiting the Earth). Distance between the Earth and

the Moon is 3.85 x 108 m. Radius of the Moon is 1.74 x 106m.

Solution:

Data: Distance between Earth and Moon = r = 3.85 x 108 m

Radius of Moon = R = 1.74 x 106 m

To find: Ration of spin and orbital angular momentum = Ls/Lo = ?

The spin angular momentum of the Moon about its axis is

sL =Iω

When

22I= MR

5

Thus,

2

s

2L = MR ω.................(1)

5

The orbital angular momentum is given by

2

oL =Mr ω.................(2)

Where angular speed in both cases is the same and also the time to complete one rotation around

Earth and one rotation around its axis is the same.

Dividing equation (1) by equation (2) we get

2 2

s

2 2

o

L 2MR ω 2R= = .................(3)

L 5Mr ω 5r




6 2

s

8 2

o

12 12

17 17

-6

L 2×(1.74×10 )=

L 5×(3.85×10 )

2×3.03×10 6.06×10= =

5×1.48×10 7.4×10

=8.2×10

-6s

o

L= 8.2×10 Ans.

L

4.1 The Earth rotates on its axis once a day. Suppose, by some process the earth contracts so

that its radius is only half as large as at present. How fast will it be rotating then? (For

sphere I = 2/5 MR2).

Solution:

Data: Time period = T1 = 24 hours

Moment of inertia of sphere = I1 = 2

12MR

5

Radius of earth 12

RR =

2

To find: How fast will the earth be rotating i.e. T2 = ?

Calculation:

According to the law of conservation of momentum

1 1 2 2I ω =I ω

Or 1 2

2 1

ω I= ...............(1)

ω I

For sphere, I1 = 2/5 Mr12

And

2 2

2 12 2

2MR (R )2MI = =

5 5 (2)

As we have,

1 2

1 2

2π 2πω = andω =

T T

Putting these values in equation, (1) we get

2

1 1

2

2 1

2π/T 2/5M(R /2)=

2π/T 2/5MR

Or T2/T1 = 1

4

Or T2 = 1T

4

As we know that

T1 = 1 day = 24 hours

T2 = 24/4 = 6 hours

T2 = 6 hours Ans.

Q.5.10 What should be the orbital speed to launch a satellite in a circular orbit 900 km above the

surface of the Earth?

(Take mass of the earth as 6.0 1024 kg and its radius as 6400 km)

Solution:

Data: height above Earth surface = 900 km = 9 105 m

Radius of Earth = 6400 km = 6.4 106m

Mass of Earth = 6.0 1024 kg

Gravitational constant = G = 6.67 10-11 Nm2 /kg2



To find: Orbital velocity = v = ?

Calculation: Using the formula

Gm

v=R +h

-11 24

6 6

6.67×10 6×10v =

6.4×10 +0.9×10

v = 7.4 km/s Ans

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